Balance the following by the method of half reactions and give the correct coefficients: ___UO^2+ + ___Cr2O7^2- + ___H+ -> ___UO2^2+ + ___Cr^3+ + ___H2O

Answer

To balance the redox reaction using the half-reaction method in acidic solution: 1. Identify and balance the oxidation half-reaction: UO^2+ -> UO2^2+. Balance Oxygen by adding one H2O to the left: UO^2+ + H2O -> UO2^2+. Balance Hydrogen by adding 2H^+ to the right: UO^2+ + H2O -> UO2^2+ + 2H^+. Balance charge by adding 2e^- to the right: UO^2+ + H2O -> UO2^2+ + 2H^+ + 2e^-. 2. Identify and balance the reduction half-reaction: Cr2O7^2- -> Cr^3+. Balance Chromium by putting a 2 in front of Cr^3+: Cr2O7^2- -> 2Cr^3+. Balance Oxygen by adding 7H2O to the right: Cr2O7^2- -> 2Cr^3+ + 7H2O. Balance Hydrogen by adding 14H^+ to the left: Cr2O7^2- + 14H+ -> 2Cr^3+ + 7H2O. Balance charge (left is +12, right is +6) by adding 6e^- to the left: Cr2O7^2- + 14H^+ + 6e^- -> 2Cr^3+ + 7H2O. 3. Equalize electrons: Multiply the oxidation half-reaction by 3 so both have 6 electrons: 3UO^2+ + 3H2O -> 3UO2^2+ + 6H^+ + 6e^-. 4. Add the half-reactions and simplify: (3UO^2+ + 3H2O) + (Cr2O7^2- + 14H^+) -> (3UO2^2+ + 6H^+) + (2Cr^3+ + 7H2O). Canceling out common terms: 3UO^2+ + Cr2O7^2- + 8H^+ -> 3UO2^2+ + 2Cr^3+ + 4H2O. The coefficients are (3, 1, 8; 3, 2, 4).